Question

If \(a,b,c\) are three distinct real numbers and \[ \lim_{x\to \infty} \frac{(b-c)x^2+(c-a)x+(a-b)} {(a-b)x^2+(b-c)x+(c-a)} = \frac{1}{2} \] then \(a+2c=\)

• A. \(b\)
• B. \(2b\)
• C. \(3b\)
• D. \(4b\)

Hint:

For limits of rational functions as \(x\to\infty\), compare the highest powers of \(x\). If numerator and denominator have the same degree, the limit is the ratio of their leading coefficients.

Answer 1

The Correct Option is C

Step 1: Recall the limit of a ratio of same-degree polynomials.
For two quadratics as $x\to\infty$, $\lim=(\text{leading coeff of numerator})/(\text{leading coeff of denominator})$.
Step 2: Apply this rule.
\[\frac{b-c}{a-b}=\frac{1}{2}.\]
Step 3: Cross-multiply and rearrange.
$2(b-c)=a-b \implies 2b-2c=a-b \implies a=3b-2c$.
Step 4: Compute $a+2c$.
$a+2c=(3b-2c)+2c=3b$.
Step 5: Verify consistency.
$a-b=2(b-c)\ne 0$ since $b\ne c$, so the denominator coefficient is non-zero and the limit is valid.
Step 6: State the answer.
\[ \boxed{a+2c=3b} \]